Answer-
The smallest angle of ABC is angle C.
Solution-
The sine law-
[tex]\dfrac{\sin A}{a}=\dfrac{\sin B}{b}=\dfrac{\sin C}{c}[/tex]
As,
[tex]=8<9<10[/tex]
[tex]=AB<AC<BC[/tex]
[tex]=c<b<a[/tex]
If, [tex]\dfrac{a}{b}=\dfrac{c}{d},\ and\ b>d,\ then\ a>c[/tex]
If two fractions are equal in value and their denominator are decreasing, then the numerators must decrease in order to keep the value of the fraction same.
So, if [tex]\dfrac{\sin A}{a}=\dfrac{\sin B}{b}=\dfrac{\sin C}{c}\ and\ c<b<a[/tex], then
[tex]\sin C<\sin B<\sin A[/tex]
As we know from the properties of trigonometry, if an angle is smaller than an another angle, then their sine will also smaller than the other.
i.e if [tex]A<B[/tex], then [tex]\sin A<\sin B[/tex]
So applying the same,
if [tex]\sin C<\sin B<\sin A[/tex], then [tex]C<B<A[/tex]
Therefore, the smallest angle of ABC is angle C.
